Exams Joint Entrance Examination Mathematics Differential Equations Question 18 Question 18 of 490 Medium MCQ
Let y = y ( x ) y=y(x) y = y ( x ) be the solution curve of the differential equation
x ( x 2 + e x ) d y + ( e x ( x − 2 ) y − x 3 ) d x = 0 , x > 0 x({x}^{2}+{e}^{x})dy+({e}^{x}(x-2)y-{x}^{3})dx=0,x>0 x ( x 2 + e x ) d y + ( e x ( x − 2 ) y − x 3 ) d x = 0 , x > 0 , passing through the point ( 1 , 0 ) (1,0) ( 1 , 0 ) . Then y ( 2 ) y(2) y ( 2 ) is equal to :
Solutions · 1 Post your solution Official solution
x ( x 2 + e x ) d y + ( e x ( x − 2 ) y − x 3 ) d x = 0 d y d x + e x ( x − 2 ) x ( x 2 + e x ) y = x 3 x ( x 2 + e x ) = e ∫ ∫ e x ( x − 2 ) x ( x 2 + e x ) d x = e ∫ e x + 2 x e x + x 2 d x − ∫ 2 x d x = e ln e − x + x 2 − 2 ln x \begin{gathered} x({x}^{2}+{e}^{x})dy+({e}^{x}(x-2)y-{x}^{3})dx=0 \\ \frac{dy}{dx}+\frac{{e}^{x}(x-2)}{x({x}^{2}+{e}^{x})}y=\frac{{x}^{3}}{x({x}^{2}+{e}^{x})} \\ ={e}^{\int \int \frac{{e}^{x}(x-2)}{x({x}^{2}+{e}^{x})}dx} \\ ={e}^{\int \frac{{e}^{x}+2x}{{e}^{x}+{x}^{2}}dx-\int \frac{2}{x}dx} \\ ={e}^{\ln {e}^{-x}+{x}^{2}-2\ln x} \end{gathered} x ( x 2 + e x ) d y + ( e x ( x − 2 ) y − x 3 ) d x = 0 d x d y + x ( x 2 + e x ) e x ( x − 2 ) y = x ( x 2 + e x ) x 3 = e ∫∫ x ( x 2 + e x ) e x ( x − 2 ) d x = e ∫ e x + x 2 e x + 2 x d x − ∫ x 2 d x = e l n e − x + x 2 − 2 l n x
= e x + x 2 x 2 ∴ y ( e 2 + x 2 x 2 ) = ∫ d x + c ⇒ y ( e x + x 2 x 2 ) = x + c \begin{gathered} =\frac{{e}^{x}+{x}^{2}}{{x}^{2}} \\ \therefore \, y(\frac{{e}^{2}+{x}^{2}}{{x}^{2}})=\int dx+c \\ \Rightarrow \, y(\frac{{e}^{x}+{x}^{2}}{{x}^{2}})=x+c \end{gathered} = x 2 e x + x 2 ∴ y ( x 2 e 2 + x 2 ) = ∫ d x + c ⇒ y ( x 2 e x + x 2 ) = x + c
Also, y ( 1 ) = 0 y(1)=0 y ( 1 ) = 0
⇒ c = − 1 ∴ y ( e 2 + x 2 x 2 ) = x − 1 \begin{gathered} \Rightarrow \, c=-1 \\ \therefore \, y(\frac{{e}^{2}+{x}^{2}}{{x}^{2}})=x-1 \end{gathered} ⇒ c = − 1 ∴ y ( x 2 e 2 + x 2 ) = x − 1
Hence, y ( 2 ) = 4 e 2 + 4 y(2)=\frac{4}{{e}^{2}+4} y ( 2 ) = e 2 + 4 4 Answer the question to unlock